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2006 AMC 12B Problems/Problem 23

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Problem

Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ , $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}{$ , where $a$ and $b$ are positive integers. What is $a+b$ ?

$\mathrm{(A)}\ 85\qquad\mathrm{(B)}\ 91\qquad\mathrm{(C)}\ 108\qquad\mathrm{(D)}\ 121\qquad\mathrm{(E)}\ 127$

Solution

Using the Law of Cosines on $\triangle PBC$ , we have:

$\begin{align*}PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow...$

Using the Law of Cosines on $\triangle PAC$ , we have: $\begin{align*}PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \...$

Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$ . $\begin{align*}\sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-20s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2...$

Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$ . Thus, $a+b=85+42=\boxed{127}$ .

See also

2006 AMC 12B (Problems • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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